POJ 2196 Computer（搜索-深度优先搜索）

Computer

Problem Description

A school bought the first computer some time ago(so this computer's id is 1). During the recent years the school bought N-1 new computers. Each new computer was connected to one of settled earlier. Managers of school are anxious about slow functioning of the net and want to know the maximum distance Si for which i-th computer needs to send signal (i.e. length of cable to the most distant computer). You need to provide this information. 


Hint: the example input is corresponding to this graph. And from the graph, you can see that the computer 4 is farthest one from 1, so S1 = 3. Computer 4 and 5 are the farthest ones from 2, so S2 = 2. Computer 5 is the farthest one from 3, so S3 = 3. we also get S4 = 4, S5 = 4.

 

Input

Input file contains multiple test cases.In each case there is natural number N (N<=10000) in the first line, followed by (N-1) lines with descriptions of computers. i-th line contains two natural numbers - number of computer, to which i-th computer is connected and length of cable used for connection. Total length of cable does not exceed 10^9. Numbers in lines of input are separated by a space.

 

Output

For each case output N lines. i-th line must contain number Si for i-th computer (1<=i<=N).

 

Sample Input

51 12 13 11 1

 

Sample Output

32344

 

Author

scnu

 

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题目大意：

告诉你一棵树，问你某个点最远能到达多远？

解题思路：

先从1号点出发，找出各个点到1号点的距离，最远的那个点必然是树的主枝干上的一点。

从树的主干上的那点出发，距离最远的另一点必然是树主枝干的另一点。

接下来的答案就是每个点到两个主干点的距离取大。

解题代码：

#include <iostream>#include <cstdio>#include <climits>#include <map>#include <vector>#include <algorithm>using namespace std;const int maxn=11000;struct edge{    int u,v,w;    int next;    edge(int u0=0,int v0=0,int w0=0){ u=u0;v=v0;w=w0;}}e[maxn*2];int n,cnt,head[maxn],d[maxn],dx[maxn],dy[maxn];void initial(){    cnt=0;    for(int i=0;i<=n;i++) head[i]=-1;}void addedge(int u,int v,int w){    e[cnt]=edge(u,v,w);e[cnt].next=head[u];head[u]=cnt++;}void input(){    int x,y,w0;    for(int i=2;i<=n;i++){        scanf("%d%d",&y,&w0);        addedge(i,y,w0);        addedge(y,i,w0);    }}void dfs(int u,int fa,int dis,int *d){    for(int i=head[u];i!=-1;i=e[i].next){        int v=e[i].v,w=e[i].w;        if(v!=fa) dfs(v,u,d[v]=dis+w,d);    }}void solve(){    int x=1,y=1;    dfs(1,-1,d[1]=0,d);    for(int i=1;i<=n;i++) if(d[x]<d[i]) x=i;    dfs(x,-1,dx[x]=0,dx);    for(int i=1;i<=n;i++) if(dx[y]<dx[i]) y=i;    dfs(y,-1,dy[y]=0,dy);    for(int i=1;i<=n;i++) d[i]=max(dx[i],dy[i]);    for(int i=1;i<=n;i++) printf("%d\n",d[i]);}int main(){    while(scanf("%d",&n)!=EOF){        initial();        input();        solve();    }    return 0;}